Question

How do you show `cos(x+pi/2)+cos(x-pi/2)=0`?

Answer 1

We need to use the trig identity:
`cos(A+-B)=cosAcosB∓sinAsinB`

Using this, we get:
`cos(x+pi/2)+cos(x-pi/2)=(cosxcos(pi/2)+sinxsin(pi/2))+(cosxcos(pi/2)-sinxsin(pi/2))`

`cos(pi/2)=0`
`sin(pi/2)=1`

`cos(x+pi/2)+cos(x-pi/2)=(0cosx+1sinx)+(0cosx-1sinx)=sinx-sinx=0`